Lotto Calculations!

[Pinch Paxton]

I need the help of a math's wizard, because I am confused by something....

 

The odds of winning £10 on the Lotto are 56.7 to 1. Ok.

 

But I have been told that if you write 57 lines of numbers, no matter what they are, you can't always win £10. In fact I have been told that you need 168 lines of numbers to guaranty £10.

 

Could anyone explain the mathematics behind this in simple terms please.

 

I am trying to write a computer program to find the smallest number of lines possible. I also want someone to be able to type 6 numbers into the computer, and the computer will make the smallest number of lines using those numbers.

 

Pincho

[jordan]

I am not sure what you mean by "lines of numbers", perticularly in your last sentence.

[Pinch Paxton]

The Lotto is 6 numbers from 49. A line of numbers would be 1 set of 6 numbers.

 

Line = 6 numbers.

 

A ticket!

 

168 tickets required to Guarantee £10???

 

I thought that 57 tickets guaranteed £10 so I am confused by that.

 

What maths can be used to work this out?

 

Pincho.

[jordan]

So 6 numbers are randomly drawn from the numbers 1 through 49. Does the order in which they are drawn matter? Are the numbers replaced after being drawn? (I would asume not) I only ask because my initial guess is that there are 1.7x10^9 possibilities for lines, meaning you will have to make more than 168 lines to ensure a winning ticket.

 

For the first number drawn, you have one number and the pool has 49, giving you a 1 in 49 chance of getting the number. On the second number you have a 1 in 48 chance as the pool decreases by one. I am pretty sure your odds are multiplied in a situation like this. So the result would be:

 

49x48x47x46x45x44 which equals about 1x10^10.

 

Your odds are now 6 in 1x10^10 or 1.7x10^9

 

This is just an idea though. It makes sense to me right now so I thought I would give my input.

[Pinch Paxton]

The order does not matter. The numbers are not replaced in the drum.

 

You can choose any 6 numbers. You just have to get 3 of them right. Odds = 56.7 to 1 of 3 from 6. How many lines before you are guaranteed £10.

 

Somebody has posted 163 (Not 168) on another forum.

 

I don't know my math's signs so ^ just confuses me. I need the full explenation in +-*/ these signs. Well I suppose I could just type your maths into my computer and it will sort it out for me.

 

Pincho.

[jordan]

There seems to be enough information now. It's just too late at night for me to be thinking about this stuff. I know there is some formula for finding the number of possibilities for choosing a line of 6 numbers from 49 numbers. If someone could post relativly soon so I could get an answer that would be appreciated.

[Pinch Paxton]

I found this. It uses 42 balls instead of 49. It has the maths for the odds, but does it help with the minimum number of lines?

 

 

Consider a lottery where there are initially 42 balls, and six are drawn. What are the odds of getting exactly three numbers correct (out of the list of six you selected)?

 

Something like the following will happen:

 

The first ball is not in your list (chances are 36/42)

The second is in your list (6/41)

The third is in your list (5/40)

The fourth is not in your list (35/39)

The fifth is not in your list (34/38)

The sixth is in your list (4/37)

Overall odds of this exact sequence occurring:

 

(36.35.34.6.5.4)/(42.41.40.39.38.37) = 0.001361092504

 

 

However, there are other orders in which the balls might have come out, 19 of them in fact - the number of combinations of three items taken from six is 6! / 3! 3! = 20. Therefore the chances of getting three numbers is 20 times the above or 0.02722185008.

 

You will note that irrespective of the order of drawing, the numerator will be 36.35… for wrong balls, and 6.5… for correct ones. For example, a different order of drawing would have given (36/42)(35/41)(34/40)(6/39)/(5/38)(4/37) - but the product is exactly the same. So we can multiply the calculated chance of a single sequence by the number of different sequences (20) that will lead to the same overall result (three hits).

 

 

 

The above should be clear enough, but it is possible to write down the equation in factorial form.

 

Let

B = Total number of balls (42)

 

 

D = Number drawn = Number selected (6)

 

 

H = Number correct ('Hits') (3)

 

 

The probability is (read on only if you have no fear of equations):

 

 

 

B-D! D! B-D! D!

 

P = ---------------------------------------------

 

B-2D+H! D-H! B! H! D-H!

[wolfson]

Oh dear we had the answer at the beginning the chances of winning £10 is 1 in 1133.119/20 or 1 in 56.7, just think Inter-cooperative equations:

 

20.00 combinations of 3 (3 balls need to win £10), from 6(6!/(3!x(6-3)!), ! = factorizing.

 

These are the correct equations, they finalized these results by averaging the wins of 4million people which also concluded the Inter-cooperative equation using Snxy, this might sound a bit advanced for lotto results, but its what you use when finding possibilities. And gl all at the lotto…. lol

[Pinch Paxton]

Yes I know that the odds are 56.7 to 1, but does that mean that if you make 57 tickets for the Lotto you are guaranteed to win? I am trying to find the minimum number of lines to guarantee a win. It's a different type of mathematics.

 

That is the problem.

 

I think it has something to do with spacing the numbers by groups. So the first line could be..

 

1,7,13,19,25,31

 

The second line...

 

2,8,14,20,26,32....

 

etc...

 

It would require using the same numbers as few times as possible, I think!

 

Thanks!

 

Pincho.

 

[wolfson]

You could win on the 1st ticket or you could win on the Millionth, the theoretical law of possibilites are Inter-cooprative quantities, thus theoretically you would win when putting 57 tickets on.

[wolfson]

There is NO guarentee that you would win hence the name possibility!

[Pinch Paxton]

There is a guarantee that you would win. So far its 163 tickets that guarantee a £10 win, but I think it should be less.

 

 

 

Edit: What I mean is that you have to put all 163 tickets on in one go. £163 to win £10. You don't have to make a profit, you just have to win £10.

 

 

Pincho.

 

 

[wolfson]

NO there is NOT a guarantee at all, If you look at nomial calculation the nTH term changes at a variable rate, which relates to mr^2, if you could then change the varaible to a constant you could then guarantee a lottery win, there will never be a guarantee, it may be likley but not guarnteed!!!!

[Pinch Paxton]

What I mean is that you have to put all 163 tickets on in one go. £163 to win £10. You don't have to make a profit, you just have to win £10.

 

 

Pincho.

[wolfson]

Maybe but ALL i am saying is that there is NO guarantee that i will win £10 when i put on 163 lottery lines, it is just a possibility. And they relate to N.

[Pinch Paxton]

There is a guaranteed £10 win for 163 lotto lines, but you have to do an exact sequence of numbers, with a certain spacing of numbers apart from each other. No matter what Lotto numbers are drawn, you will win £10, because you have all cases of three numbers from any 6.

 

Pincho.

[Pinch Paxton]

Lets make this a simpler example..

 

Lets say that the Lotto had just 6 numbers, and you just have to Guarantee 2 numbers.

 

1-2

1-3

1-4

1-5

1-6

2-3

2-4

2-5

2-6

3-4

3-5

3-6

4-5

4-6

 

14 tickets guarantee 2 right.

 

Pincho.

[jordan]

Your last post was on the right track, Pinch. What you need is to find how many possiblilities there are for picking three numbers from 42. You found 14 for picking two numbers from six (I think you missed 5-6, making the total 15 tickets). This would take a lot longer for three numbers from 42, however. (Mabey one of your previous posts covered an equation for this, I don't have a lot of time to check now) When you do that, it seems like you could devide your total by 2 as a ticket is 6 numbers, or 2 sets of 3 numbers. By doing this you have covered every possibility the machine can draw. You certainly won't make a profit though.

[jordan]

New idea....

 

Deviding by 2 wouldn't be enough. By placing three sets together, you have made more than two different sets. I came up with 20.

 

A ticket with numbers 1-2-3-4-5-6 would not only have 1-2-3 and 4-5-6 covered like I origionaly thought.

It would cover

1-2-4

1-2-5

1-2-6 etc.

 

I got 20 sets of three on each ticket, although I'm not sure if there is overlapping or not. If each ticket does have 20 sets of three, you 163 looks a lot more reasonable.

[Pinch Paxton]

Ahhh yes I remember reading that there are 20 sets of 3 numbers somewhere.

 

Yes you're right I missed 5-6. Sorry that one of my examples confused you. It was from the American Lotto 6/42. My calculations are from the UK Lotto 6/49. I'm still stuck!

 

Pincho.

[Emmon]

i keep getting 110,544 for the 49 version ( one to win a tenner )

 

if its 49 balls and you need 3 i work it like this;

 

You need 1 number so thats 1/49 in getting that.

 

Then you need another so thats 1/48 ( as youve already got 1 number )

 

then another, so thats 1/47.

 

49 x 48 x 47 = 110,544

 

where do you get 134 from?

[wolfson]

Your odds are 1 in 133.5 chance is 7.49E-3 (0.00749), the equation I used is M1(x) = Si(xi-E(x))nP(xi)

=SixiP(xi)-Sie(x)P(xi)= E(X)-E(X)SiP(Xi)=0. Though this would not be worth doing as spending £134 on lotto tickets to win £10 is just not ethical. The equation goes on a bit more till you transpose to reach N^2 which ends as E(X^2)-EP(X^2).

[Pinch Paxton]

Sounds good! can anyone please varify this? Can anyone suggest the............

 

For Next loop

 

in programming to create the lines?

 

Thanks!

 

Pincho.

[wolfson]

 

n_C_k = n!/ k!(n - k)! !=Fractionating

 

This is a simpliar way of writing it out for you, i used the from the final transposed sections of n^2

[jordan]

Emmon, you are assuming that the order matters. If it did, each ball (now changed to 42 of them) would have the odds you said, but since order doesn't matter, the odds are less.

 

The way I see it now is this,

 

The machine can pick any 3 numbers it wants. To find the number of possibilities, you use the equation:

 

n!

------

r!((n-r)!)

 

were n=number of balls (42) and r=number drawn(3)

 

The answer I get is 11480. As I said earlier, joining two sets of three on a ticket of six yeilds 20 sets of three. So deviding the number of sets that can be drawn (11480) by the number each ticket holds (20) means that you would need 574 tickets to guarentee a win.

 

wolfson, I didn't understand your last two post. If you could explain them to me (where you got the equations and what the variables are) that would be helpful as I am determined to get an answer now.